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Question

Calculate the energy released when 8000 small water droplets, each of same radius 2×107 m coalesce to form one large drop. The surface tension of water is 7.5×102 N/m

A
4.2×1010 J
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B
2.8×1010 J
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C
zero
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D
6.5×1010 J
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Solution

The correct option is B 2.8×1010 J
Let r be the radius of smaller drops and R of the bigger one.
Equating the initial and final volume, we have
43πR3=8000×(43πr3)
R3=8000r3
R=(8000)13r
R=20r=20×2×107=4×106 m

Increase in surface area ΔA=4πR28000(4πr2)
=4π×400r28000×4πr2
=7600×4π×r2
=7600×4π×4×1014
[-ve sign shows surface area is decreasing ]
Energy released in the process
U=T|ΔA|=7.5×102×7600×16π×1014
=2.8×1010 J

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